Question

Saturn's moon Mimas has an orbital period of 82,800 s at a distance of 1.87x10^8m from Saturn. Using m central m= (4n^2d^3/GT^2) 1 determine Saturn's mass.
Determine Saturn's mass by rearranging Newton's version of Kepler's Third Law.

Answer 1

Answer:

$5.65\times 10^{26} kg$

Explanation:

From Kepler's third law: Mass of the planet is given by:

$M = \frac{4\pi ^2d^3}{GT^2}$

where, T is the time period of satellite revolving about the planet at a distance d. G is the gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²

Given, d = 1.87 × 10⁸ m

T = 82800 s

⇒$M = \frac{4\pi ^2 (1.87\times 10^8)^3}{(6.67\times 10^{-11})(82800)^2}=5.65\times 10^{26} kg$

Answer 2

5.65×10^26kg here you go